Let $G$ be a group, and let $(A_{i})_{{i\in I}}$ be a family of subgroups of $G$. Then $G$ is said to be a free product of the subgroups $A_{i}$ if given any group $H$ and a homomorphism $f_{i}\colon A_{i}\to H$ for each $i\in I$, there is a unique homomorphism $f\colon G\to H$ such that $f|_{{A_{i}}}=f_{i}$ for all $i\in I$. The subgroups $A_{i}$ are then called the free factors of $G$.

If $G$ is the free product of $(A_{i})_{{i\in I}}$, and $(K_{i})_{{i\in I}}$ is a family of groups such that $K_{i}\cong A_{i}$ for each $i\in I$, then we may also say that $G$ is the free product of $(K_{i})_{{i\in I}}$. With this definition, every family of groups has a free product, and the free product is unique up to isomorphism.

The free product is the coproduct in the category of groups.

Free groups are simply the free products of infinite cyclic groups, and it is possible to generalize the construction given in the free group article to the case of arbitrary free products. But we will instead construct the free product as a quotient of a free group.

Let $(K_{i})_{{i\in I}}$ be a family of groups. For each $i\in I$, let $X_{i}$ be a set and $\gamma_{i}\colon X_{i}\to K_{i}$ a function such that $\gamma_{i}(X_{i})$ generates $K_{i}$. The $X_{i}$ should be chosen to be pairwise disjoint; for example, we could take $X_{i}=K_{i}\times\{i\}$, and let $\gamma_{i}$ be the obvious bijection. Let $F$ be a free group freely generated by $\bigcup_{{i\in I}}X_{i}$. For each $i\in I$, the subgroup ${\langle X_{i}\rangle}$ of $F$ is freely generated by $X_{i}$, so there is a homomorphism $\phi_{i}\colon{\langle X_{i}\rangle}\to K_{i}$ extending $\gamma_{i}$. Let $N$ be the normal closure of $\bigcup_{{i\in I}}\ker{\phi_{i}}$ in $F$.

Then it can be shown that $F/N$ is the free product of the family of subgroups $({\langle X_{i}\rangle}N/N)_{{i\in I}}$, and $K_{i}\cong{\langle X_{i}\rangle}N/N$ for each $i\in I$.